# Thread: [SOLVED] f integrable implies f^2 integrable

1. ## [SOLVED] f integrable implies f^2 integrable

I'm trying, unsuccessfully at the moment, to show the following, which I know is true: If $\displaystyle f$ is Riemann integrable, then $\displaystyle f^2$ is Riemann integrable. And, I need to do this without the u,v integrable implies uv integrable. I know there's a way to do it, so if you can point me in the right direction, that would be great! Thanks!

2. A simple example...

$\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{x}}$ exists...

$\displaystyle \int_{0}^{1} \frac{dx}{x}$ doesn't exist...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.

4. Can you use that if $\displaystyle g(x)$ is continuous and $\displaystyle f(x)$ integrable then $\displaystyle (g\circ f)(x)$ is integrable?

If so then just take $\displaystyle g(x)=x^2$.

5. Yessir, that's the trick. The composition of an integrable function into a continuous function maintains integrability. So, yea, taking your continuous function to be $\displaystyle t^2$ or whatever gives that$\displaystyle f\in R[a,b]\Rightarrow f^2\in R[a,b]$
Cheers!

6. Originally Posted by nqramjets
f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.
I'm afraid I cant agree with you. Starting from the general integration formula...

$\displaystyle \int x^{a}\cdot dx = \frac{x^{1+a}}{1+a} + c$

... where a is an arbitrary real [or even complex...] number provided only that $\displaystyle a \ne -1$, proceeding in trivial way we have...

$\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{x}} = |2\cdot \sqrt{x}|_{0}^{1}= 2$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. Okay, this may be a slight difference of definition, but here is the issue:
A function is $\displaystyle f:[0,1]\rightarrow \mathbb{R}$ is Riemann Integrable on $\displaystyle [0,1]$ if there is an $\displaystyle L\in \mathbb{R}$ such that for all $\displaystyle \varepsilon >0$ there exists $\displaystyle \delta >0$ such that for every tagged partition $\displaystyle \dot{P}$ of $\displaystyle [0,1]$ with $\displaystyle \|\dot{P}\|<\delta$, then
$\displaystyle |S(f;\dot{P})-L|<\varepsilon$
Now, if you look at the Riemann sum part you have
$\displaystyle S(f;\dot{P})=\sum_{i=1}^nf(t_i)(x_i-x_{i-1})$ but if I choose my tag of the first sub-interval to be 0, ie. $\displaystyle t_1=0\in [x_0,x_1]$ then $\displaystyle f(t_i)=\frac{1}{\sqrt{t_i}}=\frac{1}{0}$ which is undefined. So, $\displaystyle f\not\in R[0,1]$ and therefore we cannot apply the FTC.
$\displaystyle \frac{1}{\sqrt{x}}$ simply is not a function from $\displaystyle [0,1]$ into $\displaystyle \mathbb{R}$ because 0 is not in the domain.

8. Originally Posted by nqramjets
Okay, this may be a slight difference of definition, but here is the issue:
A function is $\displaystyle f:[0,1]\rightarrow \mathbb{R}$ is Riemann Integrable on $\displaystyle [0,1]$ if there is an $\displaystyle L\in \mathbb{R}$ such that for all $\displaystyle \varepsilon >0$ there exists $\displaystyle \delta >0$ such that for every tagged partition $\displaystyle \dot{P}$ of $\displaystyle [0,1]$ with $\displaystyle \|\dot{P}\|<\delta$, then
$\displaystyle |S(f;\dot{P})-L|<\varepsilon$
Now, if you look at the Riemann sum part you have
$\displaystyle S(f;\dot{P})=\sum_{i=1}^nf(t_i)(x_i-x_{i-1})$ but if I choose my tag of the first sub-interval to be 0, ie. $\displaystyle t_1=0\in [x_0,x_1]$ then $\displaystyle f(t_i)=\frac{1}{\sqrt{t_i}}=\frac{1}{0}$ which is undefined. So, $\displaystyle f\not\in R[0,1]$ and therefore we cannot apply the FTC.
$\displaystyle \frac{1}{\sqrt{x}}$ simply is not a function from $\displaystyle [0,1]$ into $\displaystyle \mathbb{R}$ because 0 is not in the domain.
I would also have to agree, since for a function to be Riemann integrable on an interval it has to be bounded. Which $\displaystyle \frac{1}{\sqrt{x}}$ clearly isn't on [0,1]. So even though $\displaystyle \int_0^1\frac{dx}{\sqrt{x}}$ exists, it is actually found by doing the following: $\displaystyle \lim_{c\to 0^+}\int_c^1\frac{dx}{\sqrt{x}}$.

9. Exactly, By definition, 1/sqrt{x} is not Riemann integrable, so the theorem that show f-->f^2 does not apply. Hence this does not break the rules. After all, a limit of an integral is a limit of a limiting process, which is where everything gets alittle funny. So, in conclusion, we are both right: f integrable implies f^2 integrable, but improper integrals do not necessarily obey the same rules.

10. $\displaystyle \chi\sigma$ has confused problems from integration theory with an improper integral from basic calculus.
In basic calculus, $\displaystyle \int_0^1 {\frac{{dx}}{{\sqrt x }}}$ exist and equals $\displaystyle 2$.
But it is also known as an improper integral because $\displaystyle \frac{1}{\sqrt x}$ is not defined at $\displaystyle x=0$.

In integration theory the function $\displaystyle \frac{1}{\sqrt x}$ is not Riemann integrable on $\displaystyle [0,1]$, not being bounded there.
But it does have an improper integral there.

11. Originally Posted by Plato
$\displaystyle \chi\sigma$ has confused problems from integration theory with an improper integral from basic calculus.
In basic calculus, $\displaystyle \int_0^1 {\frac{{dx}}{{\sqrt x }}}$ exist and equals $\displaystyle 2$.
But it is also known as an improper integral because $\displaystyle \frac{1}{\sqrt x}$ is not defined at $\displaystyle x=0$.

In integration theory the function $\displaystyle \frac{1}{\sqrt x}$ is not Riemann integrable on $\displaystyle [0,1]$, not being bounded there.
But it does have an improper integral there.
I believe it is Lebesgue integral though.

12. Yes, I'm pretty sure it is.

13. The original Riemann definition is reported here...

http://en.wikipedia.org/wiki/Riemann_integral

... and here we can read...

Given an f(*) defined in the closed interval $\displaystyle [a,b]$, the finite sequence $\displaystyle a = x_{0}<x_{1}<...<x_{n} = b$, the finite sequence $\displaystyle t_{0},t_{1},...,t_{n-1}$ subject to the condition $\displaystyle x_{i}\le t_{i} \le x_{i+1}$, a real number $\displaystyle \epsilon >0$, if exists a real number $\displaystyle \delta>0$ so that for any partition $\displaystyle \overrightarrow{x},\overrightarrow{t}$ is...

$\displaystyle |\sum_{i=0}^{n-1} f(t_{i})\cdot (x_{i+1}-x_{i}) -s|<\epsilon$ (1)

... we define...

$\displaystyle s=\int_{a}^{b} f(x)\cdot dx$ (2)

Now I have indicated as example the function...

$\displaystyle f(x) = \frac{1}{\sqrt{x}}$ , $\displaystyle 0<x\le 1$ (3)

It is obvious that the (3) is not defined in $\displaystyle x=0$ , but, as in most of such cases, we can resolve the abiguity defining f(*) [for example...] as follows...

$\displaystyle f(x)= \frac{1}{\sqrt{x}}$ for $\displaystyle 0<x\le 1$, $\displaystyle =0$ for $\displaystyle x=0$ (4)

... so that is...

$\displaystyle f^{2}(x)= \frac{1}{x}$ for $\displaystyle 0<x\le 1$, $\displaystyle =0$ for $\displaystyle x=0$ (5)

Now we have to verify if $\displaystyle f(*)$ defined in (4) is Riemann integrable according to (1) and if $\displaystyle f^{2}(*)$ defined in (5) is not Riemann integrable according to (1)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

14. Now we try to compute the 'Riemann sum' ...

$\displaystyle \sum_{i=0}^{n-1} f(t_{i})\cdot (x_{i+1} - x_{i})$ (1)

... having as goal to arrive to the integral...

$\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{x}} = 2$ (2)

The most simple choice is...

$\displaystyle x_{i}= \frac {i}{n}$

$\displaystyle t_{i}= 0$ for $\displaystyle i=0$ , $\displaystyle t_{i}=\frac{i+\alpha}{n}$ for $\displaystyle i=1,2,...,n-1$ with $\displaystyle 0<\alpha<1$ (3)

Defining...

$\displaystyle f(x)= \frac{1}{\sqrt{x}}= 0$ in $\displaystyle x=0$ (4)

... the integral (2) is...

$\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{x}} = lim_{n \rightarrow \infty} \sum_{i=1}^{n-1} \frac {f(t_{i})}{n} = lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}\cdot \sum _{i=1}^{n-1} \frac{1}{\sqrt{i +\alpha}}$ (5)

The limit (5) has been computed in elegant fashion in ...

http://www.mathhelpforum.com/math-he...-sequence.html

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

15. Obviously if I want to compute the integral of 1/sqrt{x} I can, it can be done is second quarter calculus. The point was that 1/sqrt{x} is NOT Reimann integrable in the strictest, by definition, sense. As a counterexample to f integrable --> f^2 integrable you MUST start with a function that is Reimann integrable. 1/sqrt{x} simply is not on [0,1]. It doesn't matter how you alter the function so that it becomes defined, for it is obviously then not the same function. So the implication is ambiguous.

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